Question

A student is running a titration in the laboratory to determine the alkalinity (Total, bicarbonate and carbonate, hydroxide) of a drinking water sample. The initial pH of the sample is 11.6; 0.02N H2SO4 is used as the titrant; and the sample volume subjected to the titration is 100 mL (0.1L). The results of the titrations were as follows: 18.3 (volume of titrant to drop the pH from 11.6 to 8.3) = 10.0 mL; V4.5 (volume of titrant to drop the pH from 8.3 to 4.5) = 8.0 mL. Determine all the alkalinity species expressed in "mg/L as CaCO3. (25 points) Determine the "Approximate Alkalinity" of the water sample above , in mg/L as CaCO3 (10 points).

Answer 1

Given = 100 mL. - 0.02N. = 10 ml. (10+8) = 18 ml. volume of water Normalily of lL SOL volume of acid uplo Phenolphthalein I end point volume of acid up to melkyl orange - end point Methyl orange alkalinity or total alkalinity (in terms of CaCO₃). au Norm des 50%. volume of y of of x y soy X1000 mg/L acid Volume of o Water Cod : 50 x 18 xD.0 2 x1000 - M-alkalinity = 1 nii loo 1 180mg/L : 500 x18x0.02 : 180 mg IL

similarly Normalily ot x1000 volume of Phenolphthalein 50 X H₂SO4 acid alkalinity mol (in terms of cacoz) o volume of water . : 50 X 0.02 x 1000 10 X TQQ (p-alkalinity ... - 50x2 = 100 mg IL Now M = 180 = 90 mall 2 . the alkalinity is thus H x com ..Po + 1 caza due to =O M-P s į come CO2 = 2(M-P). » COS. 2(180-100): 160 mgIns

он 8 o . - 29 но о - 2 . во – 16o . Фи он 20 mg/L H alleatining altealinity OH = 20 mell due to . 1b o mgl е н сор = 0